How do you find the limit #lim (3-sqrt3^x)/(9-3^x)# as #x->2#?

You can factorize the denominator by the rule:

Then

Let's substitute and simplify:

To find the limit of (3 - sqrt(3^x))/(9 - 3^x) as x approaches 2, we can substitute the value of x into the expression and evaluate it.

Plugging in x = 2, we get (3 - sqrt(3^2))/(9 - 3^2). Simplifying further, we have (3 - sqrt(9))/(9 - 9), which becomes (3 - 3)/(9 - 9).

This simplifies to 0/0, which is an indeterminate form. To evaluate this limit, we can use L'Hôpital's Rule.

Differentiating the numerator and denominator separately, we get (0 - 0)/(0 - 0), which is still an indeterminate form.

Applying L'Hôpital's Rule again, we differentiate once more. The numerator differentiates to 0, and the denominator differentiates to 0.

We are left with 0/0 again, so we apply L'Hôpital's Rule once more. Differentiating again, both the numerator and denominator differentiate to 0.

We now have 0/0 for the fourth time. Applying L'Hôpital's Rule again, we differentiate once more. The numerator differentiates to 0, and the denominator differentiates to 0.

After four applications of L'Hôpital's Rule, we are still left with 0/0. This suggests that we need to try a different approach.

To simplify the expression further, we can rewrite it as (3 - sqrt(3^x))/(9 - 3^x) * (3 + sqrt(3^x))/(3 + sqrt(3^x)).

Multiplying the numerator and denominator by the conjugate of the numerator, we get [(3 - sqrt(3^x))(3 + sqrt(3^x))]/[(9 - 3^x)(3 + sqrt(3^x))].

Expanding the numerator, we have (9 - (sqrt(3^x))^2)/(9 - 3^x)(3 + sqrt(3^x)).

Simplifying further, we get (9 - 3^x)/(9 - 3^x)(3 + sqrt(3^x)).

Canceling out the common factors of (9 - 3^x), we are left with 1/(3 + sqrt(3^x)).

Now, we can substitute x = 2 into the expression. Plugging in x = 2, we have 1/(3 + sqrt(3^2)), which simplifies to 1/(3 + sqrt(9)).

Further simplifying, we get 1/(3 + 3), which becomes 1/6.

Therefore, the limit of (3 - sqrt(3^x))/(9 - 3^x) as x approaches 2 is 1/6.